Higher Engineering Mathematics

(Greg DeLong) #1
352 DIFFERENTIAL CALCULUS

Similarly,

∂b
∂y

=

y

(x^2 +y^2 +z^2 )

and

∂b
∂z

=

z

(x^2 +y^2 +z^2 )
dx
dt

=6 mm/s= 0 .6 cm/s,

dy
dt

=5 mm/s= 0 .5 cm/s,

and

dz
dt

=4 mm/s= 0 .4 cm/s

Hence

db
dt

=

[
x

(x^2 +y^2 +z^2 )

]

(0.6)

+

[
y

(x^2 +y^2 +z^2 )

]

(0.5)

+

[
z

(x^2 +y^2 +z^2 )

]

(0.4)

Whenx=5 cm,y=4 cm andz=3 cm, then:


db
dt

=

[
5

(5^2 + 42 + 32 )

]

(0.6)

+

[
4

(5^2 + 42 + 32 )

]

(0.5)

+

[
3

(5^2 + 42 + 32 )

]

(0.4)

= 0. 4243 + 0. 2828 + 0. 1697 = 0 .8768 cm/s

Hence the rate of increase of diagonalAC is
0.88 cm/s or 8.8 mm/s, correct to 2 significant
figures.

Now try the following exercise.

Exercise 142 Further problems on rates of
change


  1. The radius of a right cylinder is increas-
    ing at a rate of 8 mm/s and the height is
    decreasing at a rate of 15 mm/s. Find the
    rate at which the volume is changing in


cm^3 /s when the radius is 40 mm and the
height is 150 mm. [+ 226 .2cm^3 /s]


  1. Ifz=f(x,y) andz= 3 x^2 y^5 , find the rate of
    change ofzwhenxis 3 units andyis 2 units
    whenxis decreasing at 5 units/s andyis
    increasing at 2.5 units/s. [2520 units/s]

  2. Find the rate of change of k, correct to
    4 significant figures, given the following
    data:k=f(a,b,c); k= 2 blna+c^2 ea; ais
    increasing at 2 cm/s; b is decreasing at
    3 cm/s;cis decreasing at 1 cm/s;a= 1 .5 cm,
    b=6 cm andc=8 cm. [515.5 cm/s]

  3. A rectangular box has sides of lengthxcm,
    ycm andzcm. Sidesxandzare expanding
    at rates of 3 mm/s and 5 mm/s respectively
    and sideyis contracting at a rate of 2 mm/s.
    Determine the rate of change of volume when
    xis 3 cm,yis 1.5 cm andzis 6 cm.
    [1.35 cm^3 /s]

  4. Find the rate of change of the total surface
    area of a right circular cone at the instant
    when the base radius is 5 cm and the height
    is 12 cm if the radius is increasing at 5 mm/s
    and the height is decreasing at 15 mm/s.
    [17.4 cm^2 /s]


35.3 Small changes


It is often useful to find an approximate value for
the change (or error) of a quantity caused by small
changes (or errors) in the variables associated with
the quantity. Ifz=f(u,v,w,...) andδu,δv,δw,...
denotesmall changes inu,v,w,...respectively,
then the corresponding approximate changeδzin
zis obtained from equation (1) by replacing the
differentials by the small changes.

Thus δz≈

∂z
∂u

δu+

∂z
∂v

δv+

∂z
∂w

δw+··· (3)

Problem 8. Pressurepand volumeVofagas
are connected by the equationpV^1.^4 =k. Deter-
mine the approximate percentage error inkwhen
the pressure is increased by 4% and the volume
is decreased by 1.5%.
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